# trim

Find trim point of dynamic system

## Syntax

```[x,u,y,dx] = trim('sys') [x,u,y,dx] = trim('sys',x0,u0,y0) [x,u,y,dx] = trim('sys',x0,u0,y0,ix,iu,iy) [x,u,y,dx] = trim('sys',x0,u0,y0,ix,iu,iy,dx0,idx) [x,u,y,dx,options] = trim('sys',x0,u0,y0,ix,iu,iy,dx0,idx,options) [x,u,y,dx,options] = trim('sys',x0,u0,y0,ix,iu,iy,dx0,idx,options,t) ```

## Description

A trim point, also known as an equilibrium point, is a point in the parameter space of a dynamic system at which the system is in a steady state. For example, a trim point of an aircraft is a setting of its controls that causes the aircraft to fly straight and level. Mathematically, a trim point is a point where the system's state derivatives equal zero. `trim` starts from an initial point and searches, using a sequential quadratic programming algorithm, until it finds the nearest trim point. You must supply the initial point implicitly or explicitly. If `trim` cannot find a trim point, it returns the point encountered in its search where the state derivatives are closest to zero in a min-max sense; that is, it returns the point that minimizes the maximum deviation from zero of the derivatives. `trim` can find trim points that meet specific input, output, or state conditions, and it can find points where a system is changing in a specified manner, that is, points where the system's state derivatives equal specific nonzero values.

`[x,u,y,dx] = trim('sys')` finds the equilibrium point of the model '`sys`', nearest to the system's initial state, `x0`. Specifically, `trim` finds the equilibrium point that minimizes the maximum absolute value of `[x-x0,u,y]`. If `trim` cannot find an equilibrium point near the system's initial state, it returns the point at which the system is nearest to equilibrium. Specifically, it returns the point that minimizes `abs(dx)` where `dx` represents the derivative of the system. You can obtain `x0` using this command.

`[sizes,x0,xstr] = sys([],[],[],0)`

`[x,u,y,dx] = trim('sys',x0,u0,y0)` finds the trim point nearest to `x0`, `u0`, `y0`, that is, the point that minimizes the maximum value of

`abs([x-x0; u-u0; y-y0])`

`[x,u,y,dx] = trim('sys',x0,u0,y0,ix,iu,iy)` finds the trim point closest to `x0`, `u0`, `y0` that satisfies a specified set of state, input, and/or output conditions. The integer vectors `ix`, `iu`, and `iy` select the values in `x0`, `u0`, and `y0` that must be satisfied. If `trim` cannot find an equilibrium point that satisfies the specified set of conditions exactly, it returns the nearest point that satisfies the conditions, namely,

```abs([x(ix)-x0(ix); u(iu)-u0(iu); y(iy)-y0(iy)]) ```

`[x,u,y,dx] = trim('sys',x0,u0,y0,ix,iu,iy,dx0,idx)` finds specific nonequilibrium points, that is, points at which the system's state derivatives have some specified nonzero value. Here, `dx0` specifies the state derivative values at the search's starting point and `idx` selects the values in `dx0` that the search must satisfy exactly.

`[x,u,y,dx,options] = trim('sys',x0,u0,y0,ix,iu,iy,dx0,idx,options)` specifies an array of optimization parameters that `trim` passes to the optimization function that it uses to find trim points. The optimization function, in turn, uses this array to control the optimization process and to return information about the process. `trim` returns the `options` array at the end of the search process. By exposing the underlying optimization process in this way, `trim` allows you to monitor and fine-tune the search for trim points.

The following table describes how each element affects the search for a trim point. Array elements 1, 2, 3, 4, and 10 are particularly useful for finding trim points.

No.DefaultDescription
10Specifies display options. 0 specifies no display; 1 specifies tabular output; -1 suppresses warning messages.
210–4Precision the computed trim point must attain to terminate the search.
310–4Precision the trim search goal function must attain to terminate the search.
410–6Precision the state derivatives must attain to terminate the search.
5N/ANot used.
6N/ANot used.
7N/AUsed internally.
8N/AReturns the value of the trim search goal function (λ in goal attainment).
9N/ANot used.
10N/AReturns the number of iterations used to find a trim point.
11N/AReturns the number of function gradient evaluations.
120Not used.
130Number of equality constraints.
14100*(Number of variables)Maximum number of function evaluations to use to find a trim point.
15N/ANot used.
1610–8Used internally.
170.1Used internally.
18N/AReturns the step length.

`[x,u,y,dx,options] = trim('sys',x0,u0,y0,ix,iu,iy,dx0,idx,options,t)` sets the time to `t` if the system is dependent on time.

Note

If you fix any of the state, input or output values, `trim` uses the unspecified free variables to derive the solution that satisfies these constraints.

## Examples

Consider a linear state-space system modeled using a State-Space block

`$\begin{array}{l}\stackrel{˙}{x}=Ax+Bu\\ y=Cx+Du\end{array}$`

The A, B, C, and D matrices to enter at the command line or in the block parameters dialog are:.

```A = [-0.09 -0.01; 1 0]; B = [ 0 -7; 0 -2]; C = [ 0 2; 1 -5]; D = [-3 0; 1 0]; ```

### Example 1

To find an equilibrium point in this model called `sys`, use:

```[x,u,y,dx,options] = trim('sys') x = 0 0 u = 0 0 y = 0 0 dx = 0 0 ```

The number of iterations taken is:

```options(10) ans = 7 ```

### Example 2

To find an equilibrium point near `x = [1;1], u = [1;1]`, enter

```x0 = [1;1]; u0 = [1;1]; [x,u,y,dx,options] = trim('sys', x0, u0); x = 1.0e-13 * -0.5160 -0.5169 u = 0.3333 0.0000 y = -1.0000 0.3333 dx = 1.0e-12 * 0.1979 0.0035 ```

The number of iterations taken is

```options(10) ans = 25 ```

### Example 3

To find an equilibrium point with the outputs fixed to `1`, use:

```y = [1;1]; iy = [1;2]; [x,u,y,dx] = trim('sys', [], [], y, [], [], iy) x = 0.0009 -0.3075 u = -0.5383 0.0004 y = 1.0000 1.0000 dx = 1.0e-15 * -0.0170 0.1483 ```

### Example 4

To find an equilibrium point with the outputs fixed to 1 and the derivatives set to 0 and 1, use

```y = [1;1]; iy = [1;2]; dx = [0;1]; idx = [1;2]; [x,u,y,dx,options] = trim('sys',[],[],y,[],[],iy,dx,idx) x = 0.9752 -0.0827 u = -0.3884 -0.0124 y = 1.0000 1.0000 dx = 0.0000 1.0000 ```

The number of iterations taken is

```options(10) ans = 13 ```

## Limitations

The trim point found by `trim` starting from any given initial point is only a local value. Other, more suitable trim points may exist. Thus, if you want to find the most suitable trim point for a particular application, it is important to try a number of initial guesses for `x`, `u`, and `y`.

## Algorithms

`trim` uses a sequential quadratic programming algorithm to find trim points. See Sequential Quadratic Programming (SQP) (Optimization Toolbox) for a description of this algorithm.

Introduced before R2006a